package com.localking.algorithm.leetcode.array

/**
 * Given an array nums and a value val, remove all instances of that value in-place and return the new length.
 * Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
 * The order of elements can be changed. It doesn't matter what you leave beyond the new length.
 *
 * Example 1:
 * Given nums = [3, 2, 2, 3], val = 3
 * Your function should return length = 2, with the first two elements of nums being 2.
 * It doesn't matter what you leave beyond the returned length.
 *
 * Example 2:
 * Given nums = [0, 1, 2, 2, 3, 0, 4, 2], val = 2
 * Your function should return length = 5, with the first five elements of numbers containing 0, 1, 3, 0, and 4.
 * Note that the order of those five elements can be arbitrary.
 * It doesn't matter what values are set beyond the returned length.
 *
 * Clarification:
 * Confused why the returned value is an integer but you answer is an array?
 * Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
 * internally you can think of this.
 * // nums is passed in by reference. (i.e., without making a copy)
 * int len = removeElement(nums, val)
 * // any modification to nums in your function would be known by the caller.
 * // using the length returned by you function. it prints the first len elements.
 * for(int i = 0; i < len; i++) {
 * print(nums[i]);
 * }
 *
 * @author jinbo
 */
object RemoveElements {
  def main(args: Array[String]): Unit = {
    val nums = Array(0, 1, 2, 2, 3, 0, 4, 2)
    val value = 3
    println(removeElement(nums, value))
  }

  def removeElement(nums: Array[Int], `val`: Int): Int = {
    var k = 0
    for (i <- nums.indices) {
      if (nums(i) != `val`) {
        nums(k) = nums(i)
        k += 1
      }
    }
    k
  }
}
